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Proof step 7 from proof size-exgr (previous step)

#(s1 ++ a) < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
#(s1 ++ a) < # s0, a ∈ s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1
Proof step 26:
#(s1 ++ a) < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ a ∈ s0, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1

Rule argument

fma: a ∈ s0

Additional information

Used formulas: a ∈ s0